# What is Implicit Differentiation for a Circle?

• MPZ
In summary, the circle has an equation of ##(x-a)^2+(y-b)^2=r^2##. It can be differentiated to get ##2xdx+2ydy=0##, which can be used to find the equation of multiple normal lines at different points.
MPZ

## Homework Statement

Hello
I have this circle with the equation : [/B]
(x-a)^2+(y-b)^2=r^2
I want to find dy/dx for it

2. Homework Equations
(x-a)^2+(y-b)^2=r^2

## The Attempt at a Solution

I am looking on the internet and it appears that I should use what is called "Implicit differentiation", can anyone use "Implicit differentiation for this circle please, thanks!

Hi,
Keep it simple and set a=b=0 . If you differentiate ##x^2+y^2 = r^2## you get ##2xdx+2ydy=0## so ##\displaystyle {{dy\over dx}= -{x\over y}}## which can be checked easily.

PS is this homework ?

BvU said:
Hi,
Keep it simple and set a=b=0 . If you differentiate ##x^2+y^2 = r^2## you get ##2xdx+2ydy=0## so ##\displaystyle {{dy\over dx}= -{x\over y}}## which can be checked easily.

PS is this homework ?
this is not a homework, i am trying to use a mathematical software to draw images using mathematical equations :) Can you please tell me a way to not "keep it simple" since i need the values of a and b since I can't draw it that small!

MPZ said:
this is not a homework
Fair enough.
MPZ said:
not "keep it simple"
The idea was to make it easier to understand, so that you would be able to complicate things on your own. Same differentiation on ##(x-a)^2+(y-b)^2 = r^2## gives you ##2(x-a)dx+2(y-b)dy=0## which is not very surprising if you see it as a translation of the origin to ##(a,b)##.

MPZ said:
draw images using mathematical equations
Can you elaborate? Give an example why you need ##dy\over dx## ?

BvU said:
Fair enough.
The idea was to make it easier to understand, so that you would be able to complicate things on your own. Same differentiation on ##(x-a)^2+(y-b)^2 = r^2## gives you ##2(x-a)dx+2(y-b)dy=0## which is not very surprising if you see it as a translation of the origin to ##(a,b)##.

Can you elaborate? Give an example why you need ##dy\over dx## ?
are you a detective? LOOL. I need the derivative because from it I can get the slope of the normal at any point since I want to find the equation of multiple normal lines at different points to draw "the hair" of the thing I am drawing. No more questions please with this sort

You hardly need calculus and derivatives for that. The normal lines to a circle are radius lines. Straight lines through the center.

cnh1995
Here's a sample written in Maple:
> restart; with(plots);
> circle := plot([cos(theta), sin(theta), theta = 0 .. 2*Pi], color = black, axes = none, thickness = 2):
> hair := seq(plot([r*cos((1/3)*Pi+(1/36)*k*Pi), r*sin((1/3)*Pi+(1/36)*k*Pi), r = 1 .. 1.2], thickness = 2), k = 0 .. 12);
> display({hair}, circle);

Maybe that will give you some ideas.

## 1. What is the equation for a circle?

The equation for a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle and r is the radius.

## 2. How do you find the derivative of a circle?

To find the derivative of a circle, you need to use implicit differentiation. This means you treat both x and y as variables and use the chain rule to find dy/dx.

## 3. What is the derivative of a circle at a specific point?

The derivative of a circle at a specific point is the slope of the tangent line at that point. It can be found by plugging the x and y coordinates of the point into the derivative equation.

## 4. Can you find the derivative of a circle at any point on the circle?

Yes, the derivative of a circle can be found at any point on the circle. However, the derivative will be different for each point.

## 5. How do you interpret the derivative of a circle?

The derivative of a circle represents the rate of change of the circle's circumference at a specific point. It can also be interpreted as the slope of the tangent line to the circle at that point.

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